Question

If a₁ (> 0), a₂, a₃, a₄, a₅ are in a G.P., a₂ + a₄ = 2a₃ + 1 and 3a₂ + a₃ = 2a₄, then a₂ + a₄ + 2a₅ is equal to _______.

Answer 1

Correct Answer: 40

 To solve the problem, we first note that since \(a_1, a_2, a_3, a_4, a_5\) are in a geometric progression (G.P.), there exists a common ratio \(r\) such that: 
\(a_2 = a_1 \cdot r,\) 
\(a_3 = a_1 \cdot r^2,\) 
\(a_4 = a_1 \cdot r^3,\) 
\(a_5 = a_1 \cdot r^4.\) 

According to the problem, we have two equations: 
1. \(a_2 + a_4 = 2a_3 + 1\) 
2. \(3a_2 + a_3 = 2a_4\) 

Let's rewrite these using the expressions for \(a_2, a_3,\) and \(a_4\): 
\(a_1r + a_1r^3 = 2a_1r^2 + 1\) or \(a_1r + a_1r^3 - 2a_1r^2 = 1\) 
Factoring out \(a_1,\) we get \(a_1(r + r^3 - 2r^2) = 1.\) 

From the second equation, substitute: 
\(3a_1r + a_1r^2 = 2a_1r^3\) or \(3r + r^2 = 2r^3\) 
Rearranging, we have \(3r + r^2 - 2r^3 = 0.\) 
Factor the equation: 
\(r(3 + r - 2r^2) = 0.\) 
Therefore, \(r = 0\) or \(3 + r - 2r^2 = 0.\) Since \(r \neq 0\) in a G.P., solve for \(r\): 
Rearrange to form a quadratic: \(2r^2 - r - 3 = 0.\) 
Using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\) where \(a = 2, b = -1, c = -3:\) 
\(r = \frac{1 \pm \sqrt{1 + 24}}{4},\) 
\(r = \frac{1 \pm 5}{4}.\) 
\(r = \frac{6}{4} = \frac{3}{2}\) or \(r = \frac{-4}{4} = -1.\) Since \(a_1 > 0\) and all terms are positive, \(r = \frac{3}{2}.\) 

With \(r = \frac{3}{2},\) substitute back to find \(a_2\): 
\(a_2 = a_1 \cdot \frac{3}{2}.\) 
Solve for \(a_1\): 
From \(a_1(r + r^3 - 2r^2) = 1,\) substitute \(r = \frac{3}{2}:\) 
\(a_1\left(\frac{3}{2} + \frac{27}{8} - \frac{18}{4}\right) = 1\) 
Convert fractions: 
\(a_1 \left(\frac{12}{8} + \frac{27}{8} - \frac{36}{8}\right) = 1\) 
\(a_1 \cdot \frac{3}{8} = 1\) 
\(a_1 = \frac{8}{3}.\) 

Calculate \(a_2, a_4,\) and \(a_5:\) 
\(a_2 = \frac{8}{3} \cdot \frac{3}{2} = 4.\) 
\(a_4 = a_1 \cdot \left(\frac{3}{2}\right)^3 = \frac{8}{3} \cdot \frac{27}{8} = 9.\) 
\(a_5 = a_1 \cdot \left(\frac{3}{2}\right)^4 = \frac{8}{3} \cdot \frac{81}{16} = \frac{27}{2}.\) 

We need to find \(a_2 + a_4 + 2a_5:\) 
\(4 + 9 + 2 \times \frac{27}{2} = 13 + 27 = 40.\) 
Verify: The computed value \(40\) is within the given range (40,40). Thus, the solution is verified:

40